Total variation metric space and its equivalence classes Let $BV[a,b]$ denote the class of all functions of bounded variation on $[a,b]$ (real interval). One can decompose $BV$ into equivalence classes according to the equivalence relation $f\sim g$ if $f(t)-g(t)$ is constant in $[a,b]$. This defines the metric space $\tilde{BV}$ with metric $\rho(f,g)=V(f-g)$, with $V(f)$ the total variation. If we define instead the metric in $BV$ as
$$
\rho(f,g) = |f(a)-g(a)| + V(f-g),
$$
why the subspace of all $f\in BV$ for which $f(a)=0$ is identified in a natural way with the space $\tilde{BV}$?
(In my definition, $V(f) = \sup \sum_{i=1}^n |f(x_i)-f(x_{i-1})|$ where the supremum is over a partition $a=x_0<x_1<\dots<x_n=b$.)
I know that $V(f-g)=0$ if and only if $f$ and $g$ differ by a constant, but I cannot see why $f(a)=0$ would imply $f(t)-g(t)$ is constant over the whole interval, or am I understanding wrong the definition of $\tilde{BV}$?
 A: It is because if you denote $X=\{f\in BV : f(a) =0 \}\subset BV $ then for $f,g\in X$ you have $[f]_{\sim }=\{h\in BV : f(x)-h(x) =\mbox{const}\} ,[g]_{\sim }=\{h\in BV : g(x)-h(x) =\mbox{const}\}$ and $$\tilde{\rho} ([f]_{\sim } , [g]_{\sim } )=\mbox{Var} (f-g)=|0-0|+\mbox{Var} (f-g)=|f(a)-g(a)|+\mbox{Var} (f-g)=\rho (f,g)$$
So the function $\xi :X\to \widetilde{BV} ,$ $\xi (f)=[f]_{\sim }$ is an linear isometry so the spaces $X$ and $\widetilde{BV}$ so the spaces can by identified.
A: The reason is simple. Take e.g. $a=0$. If two paths coincide at $0$ and they differ by a constant, then they are identical.
In more detail: Take $a$ to be arbitrary. For each path $X$ in $BV[a,b]$, one can construct an equivalence class on $BV[a,b]$ defined as the set of paths that differ from $X$ by a constant. In particular, only one such path in the equivalence class of $X$, given by $X^*_t = X_t -X_a$, is equal to $0$ at $a$. Hence, a natural way in which $\tilde{BV}$ can be identified with $BV[a,b]$ is by picking the representative of the equivalence class of each path that is equal to $0$ at $a$.
